Re: [algogeeks] Please explain the output
#a is the replacement sequence which is substituted in the printf statement
The statements
#define power(a) #a
printf("%d",power(a));
is substituted as
printf("%d","a");
it is replaced with the string literal "a" . then *power(a) is converted as
value at that string literal address.
Hope this solves the problem :)
--
Rajeev N B
I Blog @ www.opensourcemania.co.cc
On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla wrote:
> #include
> #define power(a) #a
> int main()
> {
> printf("%d",*power(432));
> return 0;
> }
>
>
> ans is 52 on gcc. Explain plss
>
> --
> best wishes!!
> Vaibhav Shukla
> DU-MCA
>
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Re: [algogeeks] Please explain the output
hmm i got it.thnx
On Thu, Jun 23, 2011 at 4:05 PM, Piyush Sinha wrote:
> printf("%d",*power(432)) will expand as
>
> *printf("%d", *"432")*
>
> "432" represents here a string and *"432" is pointing to the first string
> literal i.e 4 whose ascii value is 52..hence the output is 52
>
>
> On Thu, Jun 23, 2011 at 4:02 PM, Shachindra A C wrote:
>
>> #include
>> #define power(a) #a
>> int main()
>> {
>> printf("%d",*power(432));
>> return 0;
>> }
>>
>> the printf statement, after preprocessing, will look like
>> printf("%d",*"432");
>>
>> so, when u print the value at the first position of the string, 52, which
>> is the ascii value of 4, will be printed.
>>
>> On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla
>> wrote:
>>
>>> #include
>>> #define power(a) #a
>>> int main()
>>> {
>>> printf("%d",*power(432));
>>> return 0;
>>> }
>>>
>>>
>>> ans is 52 on gcc. Explain plss
>>>
>>> --
>>> best wishes!!
>>> Vaibhav Shukla
>>> DU-MCA
>>>
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>>
>>
>>
>> --
>> Regards,
>> Shachindra A C
>>
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best wishes!!
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Re: [algogeeks] Please explain the output
printf("%d",*power(432)) will expand as
*printf("%d", *"432")*
"432" represents here a string and *"432" is pointing to the first string
literal i.e 4 whose ascii value is 52..hence the output is 52
On Thu, Jun 23, 2011 at 4:02 PM, Shachindra A C wrote:
> #include
> #define power(a) #a
> int main()
> {
> printf("%d",*power(432));
> return 0;
> }
>
> the printf statement, after preprocessing, will look like
> printf("%d",*"432");
>
> so, when u print the value at the first position of the string, 52, which
> is the ascii value of 4, will be printed.
>
> On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla
> wrote:
>
>> #include
>> #define power(a) #a
>> int main()
>> {
>> printf("%d",*power(432));
>> return 0;
>> }
>>
>>
>> ans is 52 on gcc. Explain plss
>>
>> --
>> best wishes!!
>> Vaibhav Shukla
>> DU-MCA
>>
>> --
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>>
>
>
>
> --
> Regards,
> Shachindra A C
>
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Re: [algogeeks] Please explain the output
#include
#define power(a) #a
int main()
{
printf("%d",*power(432));
return 0;
}
the printf statement, after preprocessing, will look like
printf("%d",*"432");
so, when u print the value at the first position of the string, 52, which is
the ascii value of 4, will be printed.
On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla wrote:
> #include
> #define power(a) #a
> int main()
> {
> printf("%d",*power(432));
> return 0;
> }
>
>
> ans is 52 on gcc. Explain plss
>
> --
> best wishes!!
> Vaibhav Shukla
> DU-MCA
>
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Re: [algogeeks] please explain the output
thx pratik
On Sat, Apr 9, 2011 at 8:13 PM, Pratik Kathalkar
wrote:
> u can see the pre-processed file using gcc -E prog_name.cand @
> bottom u can see what actually the code is doing.
>
>
> On Tue, Apr 5, 2011 at 12:45 PM, Arvind wrote:
>
>> #include
>>
>> #define f(a,b) a##b
>> #define g(a) #a
>> #define h(a) g(a)
>>
>> int main()
>> {
>> printf("%s",g(f(1,2)));
>> printf("\t%s",h(f(1,2)));
>> return 0;
>> }
>>
>>
>>
>> i have run this program in gcc compiler and getting : f(1,2) 12 as
>> output.
>> can anyone explain the reason for getting this output?
>>
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>>
>
>
> --
> Pratik Kathalkar
> CoEP
> BTech IT
> 8149198343
>
>
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Re: [algogeeks] please explain the output
u can see the pre-processed file using gcc -E prog_name.cand @
bottom u can see what actually the code is doing.
On Tue, Apr 5, 2011 at 12:45 PM, Arvind wrote:
> #include
>
> #define f(a,b) a##b
> #define g(a) #a
> #define h(a) g(a)
>
> int main()
> {
> printf("%s",g(f(1,2)));
> printf("\t%s",h(f(1,2)));
> return 0;
> }
>
>
>
> i have run this program in gcc compiler and getting : f(1,2) 12 as
> output.
> can anyone explain the reason for getting this output?
>
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Re: [algogeeks] please explain the output
nice explanation -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] please explain the output
COOL BRO THIS IS A GOOD SOLN
On Tue, Apr 5, 2011 at 4:10 PM, Azhar Hussain wrote:
> Few Important things about macros, before I explain the output
> 1. Macros are replaced in passes.
> 2. Macros are not recursive.
>
> regarding the output remember the rule for expansion
> "A parameter in the replacement list, *UNLESS* preceded by a # or ##
> preprocessing token or followed by a ## preprocessing token, is replaced by
> the corresponding argument after all macros contained therein have been
> expanded".
> In other words, macros are replaced "inside out" unless # or ## exists
>
> printf("%s",g(f(1,2))); is replaced as #f(1,2) ---> "f(1,2)" according to
> the replacement rule.
> printf("\t%s",h(f(1,2))); As this does not replace with # or ## directly,
> "inside out" expansion leads to h("1,2") --> g('1,2") --> "1,2"
>
> for the first pass
> printf("%s", "f(1,2)"); ---> g(a) #a
> printf("\t%s", h("1,2"));
>
> second pass
> printf("%s", "f(1,2)"); >> not processed(exhausted)
> printf("\t%s", g("1,2")); --> h(a) g(a)
>
> Third pass
> printf("%s", "f(1,2)"); >> not processed(exhausted)
> printf("\t%s", "1,2"); --> g(a) #a
>
>
> Hope this answers your question.
>
> -
> Azhar.
>
>
>
>
> On Tue, Apr 5, 2011 at 3:22 PM, Vandana Bachani wrote:
>
>> Hi Arvind,
>> These are preprocessor specific operators. Check out
>> http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
>>
>> -Vandana
>>
>> On Tue, Apr 5, 2011 at 12:45 PM, Arvind wrote:
>>
>>> #include
>>>
>>> #define f(a,b) a##b
>>> #define g(a) #a
>>> #define h(a) g(a)
>>>
>>> int main()
>>> {
>>> printf("%s",g(f(1,2)));
>>> printf("\t%s",h(f(1,2)));
>>> return 0;
>>> }
>>>
>>>
>>>
>>> i have run this program in gcc compiler and getting : f(1,2) 12 as
>>> output.
>>> can anyone explain the reason for getting this output?
>>>
>>> --
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Re: [algogeeks] please explain the output
Few Important things about macros, before I explain the output
1. Macros are replaced in passes.
2. Macros are not recursive.
regarding the output remember the rule for expansion
"A parameter in the replacement list, *UNLESS* preceded by a # or ##
preprocessing token or followed by a ## preprocessing token, is replaced by
the corresponding argument after all macros contained therein have been
expanded".
In other words, macros are replaced "inside out" unless # or ## exists
printf("%s",g(f(1,2))); is replaced as #f(1,2) ---> "f(1,2)" according to
the replacement rule.
printf("\t%s",h(f(1,2))); As this does not replace with # or ## directly,
"inside out" expansion leads to h("1,2") --> g('1,2") --> "1,2"
for the first pass
printf("%s", "f(1,2)"); ---> g(a) #a
printf("\t%s", h("1,2"));
second pass
printf("%s", "f(1,2)"); >> not processed(exhausted)
printf("\t%s", g("1,2")); --> h(a) g(a)
Third pass
printf("%s", "f(1,2)"); >> not processed(exhausted)
printf("\t%s", "1,2"); --> g(a) #a
Hope this answers your question.
-
Azhar.
On Tue, Apr 5, 2011 at 3:22 PM, Vandana Bachani wrote:
> Hi Arvind,
> These are preprocessor specific operators. Check out
> http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
>
> -Vandana
>
> On Tue, Apr 5, 2011 at 12:45 PM, Arvind wrote:
>
>> #include
>>
>> #define f(a,b) a##b
>> #define g(a) #a
>> #define h(a) g(a)
>>
>> int main()
>> {
>> printf("%s",g(f(1,2)));
>> printf("\t%s",h(f(1,2)));
>> return 0;
>> }
>>
>>
>>
>> i have run this program in gcc compiler and getting : f(1,2) 12 as
>> output.
>> can anyone explain the reason for getting this output?
>>
>> --
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>>
>>
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Re: [algogeeks] please explain the output
Hi Arvind,
These are preprocessor specific operators. Check out
http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
-Vandana
On Tue, Apr 5, 2011 at 12:45 PM, Arvind wrote:
> #include
>
> #define f(a,b) a##b
> #define g(a) #a
> #define h(a) g(a)
>
> int main()
> {
> printf("%s",g(f(1,2)));
> printf("\t%s",h(f(1,2)));
> return 0;
> }
>
>
>
> i have run this program in gcc compiler and getting : f(1,2) 12 as
> output.
> can anyone explain the reason for getting this output?
>
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Re: [algogeeks] please explain the output
*The prototype of printf is*
*int printf(const char ***format**, ...);*
*
*
*Thus it takes a string and then variable number of arguments.*
*
*
*Every argument passed after (char *format)string is to resolve the"%"
inside the string (which is passed as the first argument)*
*
*
*Thus "anuj" will be printed as normal as printf("anuj");*
*Since "anuj" which is char *format for the printf function does not need to
resolve % because there aren't any.*
*
*
It will be good to read the printf implementation in Dennis Ritchie to get
a better understanding
On Fri, Nov 5, 2010 at 7:58 PM, ANUJ KUMAR wrote:
> #include
> int main()
> {
>printf("anuj","kumar");
>return 0;
> }
>
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